Answer
$ f(x)=\frac{\cos x^2}{x^2-1}$ is continuous on $ R- \{-1,1\}$.
Work Step by Step
Since $ x^2-1=0$, then $ x=\pm 1$. Then, $ f(x)=\frac{\cos x^2}{x^2-1}$ is defined for all $ x\in R\backslash \{-1,1\}$. Now, since $ x^2-1$ and $\cos x^2$ are continuous, then by using the continuity law $ f(x)=\frac{\cos x^2}{x^2-1}$ is continuous on $ R- \{-1,1\}$.
